Lecture notes on Computational Fluid Dynamics

Transcription

1 Lectre notes on Comptatonal Fld Dnamcs Dan S. Hennngson Martn Berggren Janar 3, 5

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3 Contents Dervaton of the Naver-Stokes eqatons 7. Notaton Knematcs Renolds transport theorem Momentm eqaton Energ eqaton Naver-Stokes eqatons Incompressble Naver-Stokes eqatons Role of the pressre n ncompressble flow Flow phscs 9. Exact soltons Vortct and streamfncton Potental flow Bondar laers Trblent flow Fnte volme methods for ncompressble flow Fnte Volme method on arbtrar grds Fnte-volme dscretzatons of D NS Smmar of the eqatons Tme dependent flows General teraton methods for stead flows Fnte element methods for ncompressble flow 7 4. FEM for an advecton dffson problem Fnte element approxmaton The algebrac problem. Assembl An example Matrx propertes and solvablt Stablt and accrac Alternatve Elements, 3D FEM for Naver Stokes A varatonal form of the Naver Stokes eqatons Fnte-element approxmatons The algebrac problem n D Stablt The LBB condton Mass conservaton Choce of fnte elements. Accrac A Backgrond materal 95 A. Iteratve soltons to lnear sstems A. Cartesan tensor notaton A.. Orthogonal transformaton A.. Cartesan Tensors A..3 Permtaton tensor A..4 Inner prodcts, crossprodcts and determnants A..5 Second rank tensors

4 A..6 Tensor felds A..7 Gass & Stokes ntegral theorems A..8 Archmedes prncple B Spplementar materal 7 B. Sllabs B. Std qestons

5 Preface These lectre notes has evolved from a CFD corse 5C) and a Fld Mechancs corse 5C4) at the department of Mechancs and the department of Nmercal Analss and Compter Scence NADA) at KTH. Erk Stålberg and Or Levn has tped most of the L A TEXformlas and has created the electronc versons of most fgres. Stockholm, Agst 4 Dan Hennngson Martn Berggren

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7 Chapter Dervaton of the Naver-Stokes eqatons. Notaton The Naver-Stokes eqatons n vector notaton has the followng form where the veloct components are defned the nabla operator s defned as the Laplace operator s wrtten as t + ) = ρ p + ν = =, v, w) =,, 3 ) = x,, ) ) =,, z x x x 3 and the followng defntons are sed = x + + z ν knematc vscost ρ denst p pressre see fgre. for a defnton of the coordnate sstem and the veloct components. The Cartesan tensor form of the eqatons can be wrtten t + j = p + ν ρ = where the smmaton conventon s sed. Ths mples that a repeated ndex s smmed over, from to 3, as follows = Ths the frst component of the vector eqaton can be wrtten ot as 7

8 , x v,, PSfrag replacements x, x w, 3 z, x 3 Fgre.: Defnton of coordnate sstem and veloct components PSfrag replacements x, x, x z, x 3, v, w, 3 x P ˆt = ) x r x, ˆt ) x, ˆt ) x Fgre.: Partcle path. = or t = p + ν x x x 3 ρ x ) x + x + x 3 t + x + v + w z = p ) ρ x + ν x + + z }{{ }. Knematcs Lagrangan and Eler coordnates Knematcs s the descrpton of moton wthot regard to forces. We begn b consderng the moton of a fld partcle n Lagrangan coordnates, the coordnates famlar from classcal mechancs. Lagrange coordnates: ever partcle s marked and followed n flow. The ndependent varables are x ntal poston of fld partcle ˆt tme where the partcle path of P, see fgre., s r = r x, ˆt ) and the veloct of the partcle s the rate of change of the partcle poston,.e.

9 = r ˆt Note here that when x changes we consder new partcles. Instead of markng ever fld partcle t s most of the tme more convenent to se Eler coordnates. Eler coordnates: consder fxed pont n space, fld flows past pont. The ndependent varables are x space coordnates t tme Ths the fld veloct = x, t) s now consdered as a fncton of the coordnate x and tme t. The relaton between Lagrangan and Eler coordnates,.e. x, ˆt ) and x, t), s easl fond b notng that the partcle poston s expressed n fxed space coordnates x,.e. x = r x, ˆt ) at the tme Materal dervatve t = ˆt Althogh t s sall most convenent to se Eler coordnates, we stll need to consder the rate of change of qanttes followng a fld partcle. Ths leads to the followng defnton. Materal dervatve: rate of change n tme followng fld partcle expressed n Eler coordnates. Consder the qantt F followng fld partcle, where F = F L x, ˆt ) = F E x, t) = F E r x, ˆt ), t ) The rate of change of F followng a fld partcle can then be wrtten F ˆt = F E r ˆt + F E t t ˆt = F E F E + t Based on ths expresson we defne the materal dervatve D Dt as ˆt D Dt = t + = + ) t In the materal or sbstantal dervatve the frst term measres the local rate of change and the second measres the change de to the moton wth veloct. As an example we consder the acceleraton of a fld partcle n a stead convergng rver, see fgre.3. The acceleraton s defned a j = D Dt = t + j whch can be smplfed n D for statonar case to a = D Dt = t + x Note that the acceleraton even f veloct at fxed x does not change. Ths has been experenced b everone n a raft n a convergng rver. The raft whch s followng the fld s acceleratng althogh the flow feld s stead. Descrpton of deformaton Evolton of a lne element Consder the two nearb partcles n fgre.4 drng tme dˆt. The poston of P can b Talor expanson be expressed as P : r ) dˆt + dr ) dˆt = = r ) + r ) dˆt + dr ˆt ) + dr ˆt ) dˆt = x + dx + ) dˆt + d ) dˆt

10 PSfrag replacements x, x, x z, x 3, v, w, 3 x x x r x, ˆt ) x, ˆt ) P ˆt = ) > x Fgre.3: Acceleraton of fld partcles n convergng rver. PSfrag replacements x, x, x z, x 3, v, w, 3 x x x r x, ˆt ) x, ˆt ) P ˆt = ) x x d dˆt P > dx dr dˆt ) dˆt P x r dˆt ) x x 3 Fgre.4: Relatve moton of two nearl partcles.

11 where we have sed ˆt dr ) = ) r dx ˆt x j = j = dx j = d ) r dx j ˆt and where x j change of wth ntal pos. dx j dfference n ntal pos. d dfference n veloct We can transform the expresson ˆt dr ) = d n Lagrange coordnates to an eqaton for a materal lne element n Eler coordnates D Dt dr ) = d = {expand n Eler coordnates} = dr j where change n veloct wth spatal poston dr j dfference n spatal pos. of partcles Relatve moton assocated wth nvarant parts We consder the relatve moton d = dr j b dvdng where e j s the deformaton rate tensor and = ξ j + e j + e j }{{} e j n ts nvarant parts,.e. ξ j = ) j ant-smmetrc part j e j = + j ) r δ j traceless part 3 r e j = r δ j sotropc part 3 r The smmetrc part of, e j, descrbes the deformaton and s consdered n detal below, whereas the ant-smmetrc part can be wrtten n terms of the vortct ω k and s assocated wth sold bod rotaton,.e. no deformaton. The ant-smmetrc part can be wrtten

12 dx d dˆt r dˆt ) dr dˆt ) P P r x R π + dϕ R3 R x x 3 r r 3 Fgre.5: Deformaton of a cbe. [ξ j ] = x x 3 ) x ) 3 x ) x x ) x 3 3 x = {se ω =, ω k = ɛ kj } ω 3 ω = ω 3 ω ω ω x 3 x 3 ) 3 x ) 3 x Deformaton of a small cbe Consder the deformaton of the small cbe n fgre.5, where we defne R l k = Rδ kl component k of sde l r l k = R l k + k R l j }{{} d l k dt = R δ kl + ) k dt x l relatve moton of sde l deformed cbe Frst, we consder the deformaton on sde, whch can be expressed as dr = r R = rk r k R The nner prodct can be expanded as r kr k = [ + ) ) ) ] 3 dt + dt + dt R x x x = {drop qadratc terms} = R + ) dt x

13 We have dropped the qadratc terms snce we are assmng that dt s small. dr becomes whch mples that dr = R + dt R = R + ) dt +... R x x = R x dt = Re dt dr dt = Re Ths the deformaton rate of sde depends on e, both traceless and sotropc part of. Second, we consder the deformaton of the angle between sde and sde. Ths can be expressed as cos π + dφ ) ) / = r r r r = r krk r mr m rnr n = δ k + ) k dt δ k + ) k dt + x x = dt + ) dt ) dt ) dt x x x x = + ) dt = e dt x x ) / dt x + x dt ) / where we have dropped qadratc terms. We se the trgonometrc dentt π ) cos + dϕ = cos π cos dϕ sn π sn dϕ dϕ whch allow s to obtan the fnal expresson dϕ dt = e Ths the deformaton rate of angle between sde and sde depends onl on traceless part of. Thrd, we consder the deformaton of the volme of the cbe. Ths can be expressed as dv = r r r 3 R 3 + = R 3 x dt x dt x 3 dt x dt + x dt x 3 dt R 3 3 x dt 3 x dt + 3 x 3 dt = R 3 + ) dt + ) dt + ) 3 dt R 3 x x x 3 = R ) 3 dt x x x 3 = R 3 k x k dt where we have agan omtted qadratc terms. Ths we have dv dt = R3 e rr and the deformaton rate of volme of cbe or expanson rate) depends on sotropc part of. In smmar, the moton of a fld partcle wth veloct can be dvded nto the followng nvarant parts

14 R 3 R R r 3 r r π + dϕ V t) S t) S t + t) n V t + t) V t + t) V t) Fgre.6: Volme movng wth the fld. ) sold bod translaton ) ξ j = ) j = x ɛ kjω k sold bod rotaton ) e j = + j ) r δ j 3 r volme constant deformaton v) e j = r 3 δ j r volme expanson rate.3 Renolds transport theorem Volme ntegral followng wth the fld Consder the tme dervatve of a materal volme ntegral,.e. a volme ntegral where the volme s movng wth the fld. We obtan the followng expressons D Dt V t) T j dv = lm T j t + t) dv T j t) dv t t V t+ t) V t) = lm T j t + t) dv T j t + t) dv t t V t+ t) V t) + T j t + t) dv T j t) dv t V t) V t) = lm t t V t+ t) V t) T j t + t) dv + V t) T j t dv The volme n the frst ntegral on the last lne s represented n fgre.6, where a volme element descrbng the change n volme between V at tme t and t + t can be wrtten as n t = k n k t dv = k n k t ds

15 Ths mples that the volme ntegral can be converted to a srface ntegral. Ths srface ntegral can n trn be changed back to a volme ntegral b the se of Gass or Greens) theorem. We have D Dt V t) [ ] T j dv = lm T j t + t) k n k ds t St) T j = T j k n k ds + t = St) [ Tj V t) t whch s the Renolds transport theorem. Conservaton of mass V t) + x k k T j ) ] dv + V t) T j t dv dv = {Gass/Green s theorem} B the sbsttton T j ρ and the se of the Renolds transport theorem above we can derve the eqaton for the conservaton of mass. We have [ D ρ ρ dv = Dt t + ] k ρ) = x k V t) V t) Snce the volme s arbtrar, the followng mst hold for the ntegrand = ρ t }{{} + k ρ) = ρ x } k {{} t + ρ k + ρ k = Dρ x k x k }{{} Dt 3 + ρ k x k }{{} 4 where we have sed the defnton of the materal dervatve n order to smplf the expresson. The terms n the expresson can be gven the followng nterpretatons: : accmlaton of mass n fxed element : net flow rate of mass ot of element 3 : rate of denst change of materal element 4 : volme expanson rate of materal element B consderng the transport of a qantt gven per nt mass,.e. T j = ρt j, we can smplf Renolds transport theorem. The ntegrand n the theorem can then be wrtten cont. eq.) t ρt j) + k ρ t j ) = t j ρ x k t + ρt j t + t j x t j k ρ) + ρ k = ρ Dt j k x k Dt whch mples that Renolds transport theorem becomes D Dt V t) where V t) agan s a materal volme..4 Momentm eqaton Conservaton of momentm ρ t j dv = V t) ρ Dt j Dt dv The momentm eqaton s based on the prncple of conservaton of momentm,.e. that the tme rate of change of momentm n a materal regon = sm of the forces on that regon. The qanttes nvolved are: F - bod forces per nt mass R - srface forces per nt area ρ - momentm per nt volme

16 V t + t) V t) S t + t) n V t + t) n II r dˆt dr ) dˆt Rn II ) π P P x sk) {}}{ x x 3 R 3 R R r 3 r r n IIIk) dl ds Rn I ) + dϕ Fgre.7: V t) Momentm balance for fld element S t) V t + t) V t) S t + t) n n R V t + t) R Fgre.8: Srface force and nt normal. n I We can pt the momentm conservaton n ntegral form as follows D ρ dv = ρf dv + R ds Dt V t) V t) V t) Usng Renolds transport theorem ths can be wrtten ρ D Dt dv = ρf dv + V t) St) St) R ds whch s Newtons second law wrtten for a volme of fld: mass acceleraton = sm of forces. To proceed R mst be nvestgated so that the srface ntegral can be transformed to a volme ntegral. In order to do that we have to defne the stress tensor. The stress tensor Remove a fld element and replace otsde fld b srface forces as n fgre.7. Here Rn) s the srface force per nt area on srface ds wth normal n, see fgre.8. Momentm conservaton for the small fld partcle leads to Lettng dl gves Now n j = n I j = nii j ρ D Dt ds dl = ρf ) ds dl + R n I j ds + R n II j whch leads to ) ds + ) ) = R n I j ds + R n II j ds R n j ) = R n j ) k ) R n IIIk) j s k) dl mplng that a srface force on one sde of a srface s balanced b an eqal an opposte srface force at the other sde of that srface. Note that t s a general prncple that the terms proportonal to the volme of a small fld partcle approaches zero faster than the terms proportonal to the srface area of the partcle. Ths the srface forces actng on a small fld partcle has to balance, rrespectve of volme forces or acceleraton terms.

17 P n V t + t) P x x x 3 T x T T 3 T 3 Re ) = T, T, T 3 ) PSfrag replacements R 3 x, x R, x R z, x Fgre.9: Defnton of srface 3 r 3 force components on a srface wth a normal n the -drecton., r v, r x π w, + dϕ 3 Re ) = T, T, T 3 ) xv t) T xs t) V t + t) xv t) r S x t, + ˆt ) t) x, ˆt ) T P ˆt = ) n V t + t) x > x x x 3 x Fgre.: Defntonof srface force components on a srface wth a normal n the -drecton. dˆt dx We now dvde the srface d dˆt forces nto components along the coordnate drectons, as n fgres.9 and r )., wth correspondng defntons dˆt for the force components on a srface wth a normal n the 3-drecton. Ths, T j s the -component dr ) dˆt of the srface force on a srface ds wth a normal n the j-drecton. Consder a fld partcle wth P srfaces along the coordnate drectons ct b a slanted srface, as n fgre.. The areas of the srface P elements are related b x x ds j = e j n ds = n j ds x 3 where ds s the area of the R 3 slanted srface. Momentm balance reqre the srface forces to balance n the element, we have R R r 3 r = R ds T n ds T n ds T 3 n 3 ds whch mples that the total srface force R can be wrtten n terms of the components of the stress tensor T j as r π + dϕ V t) R = T n + T n + T 3 n 3 = T j n j S t) V t + t) V t) e S t + t) R n n V t + t) R ds ds ds 3 e e 3 Fgre.: Srface force balance on a fld partcle wth a slanted srface.

18 Here the dagonal components are normal stresses off-dagonal components are P ˆt shear = stresses. Frther consderaton of the moment balance arond a fld partcle show that T j s axsmmetrc tensor,.e. > T j = T x j x x 3 Momentm eqaton x Usng the defnton of the srface force n terms of the stress tensor, the momentm eqaton dˆt can now be dx wrtten d ρ D [ Dt dv = ρf dv + T j n j ds = ρf + T ] dˆt r ) dˆt j dv x dr ) dˆt j V t) V t) The volme s agan arbtrar mplng that the ntegrand mst tself eqal zero. We have x ρ D Dt = ρf + T j x x 3 R 3 R Pressre and vscos stress tensor St) For fld at rest onl normal stresses present otherwse fld element wold deform. We ths dvde the stress tensor nto an sotropc part, the hdrodnamc pressre p, and a part dependng r on the moton of the fld. We have r π + dϕ p T V t) j = pδ j + τ j S t) p - hdrodnamc pressre, drected nward V t + t) V t) τ j - vscos stress tensor, depends on fld moton S t + t) n V t + t) p Newtonan fld It s natral to assme that the vscos stresses are fnctons of deformaton rate e j or stran. Recall that the nvarant smmetrc parts are V t) e j = + j ) r δ j + r δ j 3 x r 3 x r }{{}}{{} e j e j P P R r 3 p where e j s the volme constant deformaton rate and e j s the nform rate of expanson. For an sotropc fld, the vscos stress tensor s a lnear fncton of the nvarant parts of e j,.e. where we have defned the two vscostes as τ j = λe j + µe j µ T ): dnamc vscost here T s the temperatre) λ T ): second vscost, often= For a Newtonan fld, we ths have the followng relatonshp between the vscos stress and the stran deformaton rate) whch leads to the momentm eqaton τ j = µe j = µ + j 3 ρ D Dt = p + [ µ + j 3 ) r δ j x r r δ j x r )] + ρf

19 .5 Energ eqaton The energ eqaton s a mathematcal statement whch s based on the phscal law that the rate of change of energ n materal partcle = rate that energ s receved b heat and work transfers b that partcle. We have the followng defntons ρ [ e + ] dv ρ F dv j R j ds n q ds energ of partcle, wth e the nternal energ work }{{ rate } of F on partcle force veloct work rate of R j on partcle heat loss from srface, wth q the heat flx vector, drected otward Usng Renolds transport theorem we can pt the energ conservaton n ntegral form as D Dt V t) ρ [e + ] dv = V t) ρf dv + St) [n T j j n q ] ds = V t) [ ρf + ] T j j q ) dv Compare the expresson n classcal mechancs, where the momentm eqaton s m = F and the assocated knetc energ eqaton s m d dt ) = F work rate = force veloct work = force dst. ) From the ntegral energ eqaton we obtan the total energ eqaton b the observaton that the volme s arbtrar and ths that the ntegrand tself has to be zero. We have ρ D e + ) Dt = ρf p ) + τ j j ) q The mechancal energ eqaton s fond b takng the dot prodct between the momentm eqaton and. We obtan ρ D ) Dt p τ j = ρf + Thermal energ eqaton s then fond b sbtractng the mechancal energ eqaton from the total energ eqaton,.e. ρ De Dt = p + τ j q The work of the srface forces dvdes nto vscos and pressre work as follows p ) = p p τ j j ) = τ j + j τ j where followng nterpretatons can be gven to the thermal and the mechancal terms : : thermal terms force deformaton ): heat generated b compresson and vscos dsspaton mechancal terms veloct force gradents ): gradents accelerate fld and ncrease knetc energ

20 The heat flx need to be related to the temperatre gradents wth Forers law q = κ T where κ = κ T ) s the thermal condctvt. Ths allows s to wrte the thermal energ eqaton as ρ De Dt = p + Φ + κ T ) where the postve defnte dsspaton fncton Φ s defned as [ Φ = τ j = µ e j e j ) ] k 3 x k = µ e j ) k δ j > 3 x k Alternatve form of the thermal energ eqaton can be derved sng the defnton of the enthalp We have whch gves the fnal reslt To close the sstem of eqatons we need a h = e + p ρ Dh Dt = De Dt + Dp ρ Dt p ρ ρ Dh Dt = Dp Dt + Φ + Dρ }{{} Dt ) ρ κ T ) ) thermodnamc eqaton e = e T, p) smplest case: e = c v T or h = c p T ) eqaton of state p = ρrt where c v and c p are the specfc heats at constant volme and temperatre, respectvel. At ths tme we also defne the rato of the specfc heats as γ = c p c v.6 Naver-Stokes eqatons The dervaton s now completed and we are left wth the Naver-Stokes eqatons. The are the eqaton descrbng the conservaton of mass, the eqaton descrbng the conservaton of momentm and the eqaton descrbng the conservaton of energ. We have Dρ Dt + ρ k x k = where ρ D Dt = p + τ j + ρf ρ De Dt = p + Φ + κ T )

21 Φ = τ j τ j = µ + j ) r δ j 3 x r and the thermodnamc relaton and the eqaton of state for a gas e = e T, p) p = ρrt We have 7 eqatons and 7 nknowns and therefore the necessar reqrements to obtan a solton of the sstem of eqatons. nknown eqatons ρ contnt 3 momentm 3 p energ e thermodn. T 7 gas law 7 Eqatons n conservatve form A slghtl dfferent verson of the eqatons can be fond b sng the dentt to obtan the sstem ρ Dt j Dt = t ρt j) + x k k ρt j ) ρ t + k ρ) = x k t ρ ) + ρ k ) = p + τ j + ρf x k e + ) t + [ρ k e + )] x k = ρf p ) + τ j j ) + κ T ) These are all of the form where U t + G) = J or U t + G x + G + G z = J s the vector of nknowns, U = ρ, ρ, ρ, ρ 3, ρ e + /)) T s the vector of the rght hand sdes and G ) = J =, ρf, ρf, ρf 3, ρ F ) T ρ ρ + pδ τ ρ + pδ τ ρ 3 + pδ 3 τ 3 ρ e + /) + p + κ T j τ j s the flx of mass, momentm and energ, respectvel. Ths form of the eqaton s sall termed the conservatve form of the Naver-Stokes eqatons.

22 .7 Incompressble Naver-Stokes eqatons The conservaton of mass and momentm can be wrtten Dρ Dt + ρ = ρ D Dt = p + τ j + ρf τ j = µ + j ) k δ j 3 k for ncompressble flow ρ = constant, whch from the conservaton of mass eqaton mples = mplng that a fld partcle experences no change n volme. Ths the conservaton of mass and momentm redce to t + j = p + ν + F ρ = snce the components of the vscos stress tensor can now be wrtten + j ) k δ j = = 3 x k We have also ntrodced the knematc vscost, ν, above, defned as ν = µ/ρ The ncompressble verson of the conservaton of mass and momentm eqatons are sall referred to as the ncompressble Naver-Stokes eqatons, or jst the Naver-Stokes eqatons n case t s evdent that the ncompressble lmt s assmed. The reason that the energ eqaton s not nclded n the ncompressble Naver-Stokes eqatons s that t decoples from the momentm and conservaton of mass eqatons, as we wll see below. In addton t can be worth notng that the conservaton of mass eqaton s sometmes referred to as the contnt eqaton. The ncompressble Naver-Stokes eqatons need bondar and ntal condtons n order for a solton to be possble. Bondar condtons BC) on sold srfaces are =. The are for obvos reasons sall referred to as the no slp condtons. As ntal condton IC) one needs to specf the veloct feld at the ntal tme t = ) =. The pressre feld does not need to be specfed as t can be obtaned once the veloct feld s specfed, as wll be dscssed below. Integral form of the Naver-Stokes eq. We have derved the dfferental form of the Naver-Stokes eqatons. Sometmes, for example when a fnte-volme dscretzaton s derved, t s convenent to se an ntegral form of the eqatons. An ntegral from of the contnt eqaton s fond b ntegratng the dvergence constrant over a fxed volme V F and sng the Gass theorem. We have ) dv F = n ds F = V An ntegral form of the momentm eqaton s fond b takng the tme dervatve of a fxed volme ntegral of the veloct, sbstttng the dfferental form of the Naver-Stokes eqaton, sng the contnt eqaton and Gass theorem. We have [ dv = t t dv = j ) + ] p ν dv = ρ V F V F V F [ j + ρ pδ j ν V F ] S dv = S F [ j n j + p ρ n ν ] n j ds

23 S t + t) n V t + t) p T w T w T U L T, U L Fgre.: Examples of length, veloct and temperatre scales. Note that the ntegral from of the contnt eqaton s a compatblt condton for BC n ncompressble flow. Dmensonless form It s often convenent to work wth a non-dmensonal form of the Naver-Stokes eqatons. We se the followng scales and non-dmensonal varables x = x L t = tu L = p = p = U p L x t = U L t where sgnfes a non-dmensonal varable. The length and veloct scales have to be chosen appropratel from the problem nder nvestgaton, so that the represent tpcal lengths and veloctes present. See fgre. for examples. If we ntrodce the non-dmensonal varables n the Naver-Stokes eqatons we fnd U L t + U L j x j = p p ρl x + νu L x j x j U L x = Now drop * as a sgn of non-dmensonal varables and dvde throgh wth U /L, we fnd t + j = p = We have defned the Renolds nmber Re as ρu p + ν U L Re = U L ν = U /L νu /L nertal forces vscos forces whch can be nterpreted as a measre of the nertal forces dvded b the vscos forces. The Renolds nmber s b far the sngle most mportant non-dmensonal nmber n fld mechancs. We also defne the pressre scale as p = ρu whch leaves s wth the fnal form of the non-dmensonal ncompressble Naver-Stokes eqatons as t + j = p + Re =

24 Non-dmensonal energ eqaton We stated above that the energ eqaton decople from the rest of the Naver-Stokes eqatons for ncompressble flow. Ths can be seen from a non-dmensonalzaton of the energ eqaton. We se the defnton of the enthalp h = c p T ) to wrte the thermal energ eqaton as DT ρc p Dt = Dp Dt + Φ + κ T and se the followng non-dmensonal forms of the temperatre and dsspaton fncton Ths leads to T = T T T w T Φ = L U µφ DT µ Dt = ρu L κ T U [ Dp + µc p c p T w T ) Dt + µ ] ρu L φ = Re P r T U Dp + c p T w T ) Dt }{{} + ) R Φ Ma a cptw T ) where a speed of sond n free-stream and Ma = U a s Mach nmber. We now drop and let Ma to obtan DT Dt = Re P r T where the Prandtl nmber P r = µcp κ, whch takes on a vale of.7 for ar. As the Mach nmber approaches zero, the fld behaves as an ncompressble medm, whch can be seen from the defnton of the speed of sond. We have a = γ ρα, α = ρ ρ p where α s the sothermal compressblt coeffcent. If the denst s constant, not dependng on p, we have that α and a, whch mples that Ma..8 Role of the pressre n ncompressble flow The role of the pressre n ncompressble flow s specal de to the absence of a tme dervatve n the contnt eqaton. We wll llstrate ths n two dfferent was. Artfcal compressblt Frst we dscss the artfcal compressblt verson of the eqatons, sed sometmes for solton of the stead eqatons. We defne a smplfed contnt eqaton and eqaton of state as T ρ t + = and p = βρ Ths allows s to wrte the Naver-Stokes eqatons as t + j ) = p + Re p t + β = Let = p v w e = β p + v w f = βv v p + v vw g = βw w vw p + β

25 be the vector of nknowns and ther flxes. The Naver-Stokes eqatons can then be wrtten or where the matrces above are defned A = e = C = g = t + e x + f + g z = Re D t + A x + B + C z = R D β v w β v v w v, B = f =, D = β v v w v the egenvales of A, B, C are the wave speeds of plane waves n the respectve coordnate drectons, the are,, ± ) + β, v, v, v ± ) v + β and w, w, w ± ) w + β Note that the effectve acostc or pressre wave speed β for ncompressble flow. Projecton on a dvergence free space Second we dscss the projecton of the veloct feld on a dvergence free space. We begn b the followng theorem. Theorem. An w n can nqel be decomposed nto where w = + p =, n = on..e. nto a fncton that s dvergence free and parallel to the bondar and the gradent of a fncton, here called p. Proof. We start b showng that and p s orthogonal n the L nner prodct., p ) p = dv = p) dv = p n ds = The nqeness of the decomposton can be seen b assmng that we have two dfferent decompostons and showng that the have to be eqvalent. Let The nner prodct between gves = w = ) + p) = ) + p) ) ) + p ) p )) and ) ) [ ) ) ) ) + ) ) p ) p ))] dv = ) ) ) dv Ths we have ) = ), p ) = p ) + C

26 r π + dϕ V t) S t) V t + t) V t) S t + t) n V t + t) p T, U T w L T U Gradent felds p w Dvergence free vectorfelds // to bondar Fgre.3: Projecton of a fncton on a dvergence free space. We can fnd an eqaton for a p wth above propertes b notng that p = w n wth p n = w n on has a nqe solton. Now, f we have = w p = = w p ; and = n = w n p n To sm p, to project w on dvergence free space let w = + p ) solve for p : w = p, ) let = w p p n = Ths s schematcall shown n fgre.3. Note also that w ). Appl to Naver-Stokes eqatons Let P be orthogonal projector whch maps w on dvergence free part,.e. Pw =. We then have the followng relatons w = Pw + p P =, P p = Applng the orthogonal projector to the Naver-Stokes eqatons, we have P t + p ) = P j + ) Re Now, s dvergence free and parallel to bondar, ths

27 However, P t = t P snce need not be parallel to the bondar. pressre, we have t = P j + Re }{{} w Ths we fnd an evolton eqaton wthot the The pressre term n the Naver-Stokes eqatons ensres that the rght hand sde w s dvergence free. We can fnd a Posson eqaton for the pressre b takng the dvergence of w, accordng to the reslt above. We fnd p = j + Re = j }{{} w ths the pressre satsfes ellptc Posson eqaton, whch lnks the veloct feld n the whole doman nstantaneosl. Ths can be nterpreted sch that the nformaton n ncompressble flow spreads nfntel fast,.e. we have an nfnte wave speed for pressre waves, somethng seen n the analss of the artfcal compressblt eqatons. Evolton eqaton for the dvergence It s common n several nmercal solton algorthms for the Naver-Stokes eqatons to dscard the dvergence constrant and nstead se the pressre Posson eqaton derved above as the second eqaton n the ncompressble Naver-Stokes eqatons. If ths s done we ma enconter soltons that are not dvergence free. Consder the eqaton for the evolton of the dvergence, fond b takng the dvergence of the momentm eqatons and sbstttng the Laplacan of the pressre wth the rght hand sde of the Posson eqaton. We have t ) = ) Re Ths the dvergence s not atomatcall zero, bt satsfes a heat eqaton. For the statonar case the solton obes the maxmm prncple. Ths states that the maxmm of a harmonc fncton, a fncton satsfng the Laplace eqaton, has ts maxmm on the bondar. Ths, the dvergence s zero nsde the doman, f and onl f t s zero everwhere on the bondar. Snce t s the pressre term n the Naver-Stokes eqatons ensres that the veloct s dvergence free, ths has mplcatons for the bondar condtons for the pressre Posson eqaton. A pror we have none specfed, bt the have to be chosen sch that the dvergence of the veloct feld s zero on the bondar. Ths ma be a dffclt constrant to satsf n a nmercal solton algorthm.

28

29 z, x 3, v, w, 3 x x x Chapter r x, ˆt ) x, ˆt ) P ˆt = ) x Flow phscs > x x x 3 x. Exact soltons dˆt Plane Poselle dx flow - exact solton for channel flow d The flow nsde dˆt the two-dmensonal channel, see fgre., s drven b a pressre dfference p p between r ) the nlet and dˆt the otlet of the channel. We wll assme two-dmensonal, stead flow,.e. dr ) dˆt P P x t =, 3 = and that the x flow s fll developed, meanng that effects of nlet condtons have dsappeared. We have x 3 R 3 = x ), = x ) R The bondar R condtons are = = at x = ±h. The contnt eqaton reads r 3 r + r + 3 π = x x x + dϕ 3 V t) where the frst and the last term dsappears de to the assmpton of two-dmensonal flow and that S t) the flow s fll developed. Ths mples that = C =, where the constant s seem to be zero from the V t + t) V t) bondar condtons. S t + t) The stead momentm eqatons read n V t + t) ρ + ρ = p + µ p x x x T, U T w whch, b applng L the assmptons above, redce to T U Gradent felds L w p ee vectorfelds // to bondar p > p ρ x + ρ x = p x + µ ρg p x h x Fgre.: Plane channel geometr. 9

30 Dvergence free vectorfelds // to bondar L h x x p > p p ds n n t Fgre.: Volme of fld flowng throgh srface ds n tme t = p x + µ d dx = p x ρg The momentm eqaton n the x -drecton or normal drecton) mples p = ρgx + P x ) p x = dp dx showng that the pressre gradent n the x -drecton or streamwse drecton) s onl a fncton of x. The momentm eqaton n the streamwse drecton mples Snce P x ) and x ) we have = dp dx + µ d dx d dx = µ We can ntegrate n the normal drecton to obtan dp dx = const. ndependent of x, x ) = µ dp dx x + C x + C The constants can be evalated from the bondar condtons ±h) =, gvng the parabolc veloct profle [ = h µ dp x ) ] dx h Evalatng the maxmm veloct at the center of the channel we have max = h µ dp dx > f flow s n the postve x -drecton. The veloct becomes Flow rate max = x ) h From fgre. we can evalate the flow rate as dq = n ds. Integratng over the channel we fnd Q = n ds = {channel} = S h h dx = h3 3µ dp dx = 4 3 h max

31 L h x x p > p p n ds n t p x Fgre.3: Stokes nstantaneosl plate. Wall shear stress The wall shear stress can be evalated from the veloct gradent at the wall. We have mplng that τ = τ = µ d = µ max x dx h τ wall = τ x= h = µ max h Vortct The vortct s zero n the streamwse and normal drectons,.e. ω = ω = and has the followng expresson n the x 3 -drecton or spanwse drecton) ω 3 = = d = max x x x dx h Stokes :st problem: nstantaneosl started plate Consder the nstantaneosl startng plate of nfnte horzontal extent shown n fgre.3. Ths s a tme dependent problem where plate veloct s set to = at tme t =. We assme that the veloct can be wrtten =, t) whch sng contnt and the bondar condtons mpl that normal veloct v =. The momentm eqatons redce to = p t ρ x + ν = p ρ g The normal momentm eqaton mples that p = ρg + p, where p s the atmospherc pressre at the plate srface. Ths gves s the followng dffson eqaton, ntal and bondar condtons for = ν t, ) =, t) = t >, t) = t > We look for a smlart solton,.e. we ntrodce a new dependent varable whch s a combnaton of and t sch that the partal dfferental eqaton redces to an ordnar dfferental eqaton. Let

32 fη) = η = C t b where C and b are constants to be chosen appropratel. We transform the tme and space dervatves accordng to t = η t d dη = d Cbtb dη = η d dη = d Ctb dη = C b d t dη If we sbsttte ths nto the eqaton for and se the defnton of η, we fnd an eqaton for f bt η df dη = νc t b d f dη where no explct dependence on and t can reman f a smlart solton s to exst. coeffcent of the two η-dependent terms have to be proportonal to each other, we have Ths, the bt = κνc t b where κ s a proportonalt constant. The exponents of these expressons, as well as the coeffcents n front of the t-terms have to be eqal. Ths gves b = C = κν We choose κ = and obtan the followng d f + ηf = η = νt where = dη. The bondar and ntal condtons also have to be compatble wth the smlart assmpton. We have, ) = f ) = Ths eqaton can be ntegrated twce to obtan, t) = f) =, t) = f ) = η f = C e ξ dξ + C Usng the bondar condtons and the defnton of the error-fncton, we have f = η e ξ dξ = erfη) π Ths solton s shown n fgre.4, both as a fncton of the smlart varable η and the nscaled normal coordnate. Note that the tme dependent solton s dffsng pward. From the veloct we can calclate the spanwse vortct as a fncton of the smlart varable η and the nscaled normal coordnate. Here the maxmm of the vortct s also changng n tme, from the nfnte vale at t = t s dffsng otward n tme. ω 3 = ω z = v x = πνt e η The vortct s shown n fgre.5, both as a fncton of the smlart varable η and the nscaled normal coordnate. Note agan that the tme dependent solton s dffsng pward.. Vortct and streamfncton Vortct s an mportant concept n fld dnamcs. It s related to the average anglar momentm of a fld partcle and the swrl present n the flow. However, a flow wth crclar streamlnes ma have zero vortct and a flow wth straght streamlnes ma have a non-zero vortct.

33 felds // to bondar n ds n t L h x x x η = 4νt p > p p p Dvergence free vectorfelds // to bondar n ds n t L h x x = η x 4νt p > p p η = p 4νt t.4. = η 4νt n n ds ds t n t n t Fgre x.4: Veloct above the nstantaneosl started plate. x a) U as a fncton of the smlart varable η. b) U, t) for varos tmes. p η = 4νt = η 4νt t η ω z = ωη) πνt t ω z πνt p η = 4νt = η 4νt t ω z πνt = ωη) η t ω z = ωη) πνt = ωη) Fgre.5: Vortct above the nstantaneosl started plate. a) ω z πνt/ as a fncton of the smlart varable η. b) ω z, t) for varos tmes. Vortct and crclaton The vortct s defned mathematcall as the crl of the veloct feld as In tensor notaton ths expresson becomes ω = ω = ɛ jk k Another concept closel related to the vortct s the crclaton, whch s defned as Γ = dx = t ds C In fgre.6 we show a closed crve C. Usng Stokes theorem we can transform that ntegral to one over the area S as k Γ = ɛ jk n ds = ω n ds S Ths allows s to fnd the followng relatonshp between the crclaton and vortct dγ = ω n ds, whch can also be wrtten dγ ds = ω n Ths, one nterpretaton s that the vortct s the crclaton per nt area for a srface perpendclar to the vortct vector. Example. The crclaton of an deal vortex. Let the azmthal veloct be gven as θ = C/r. We can calclate the crclaton of ths flow as π Γ = θ r dθ = C dθ = πc Ths we have the followng relatonshp for the deal vortex C θ = Γ πr C S

34 = η 4νt t η ω z = ωη) πνt t dx = t ds S C Fgre.6: Integral along a closed crve C wth enclosed area S. The deal vortex has ts name from the fact that ts vortct s zero everwhere, except for an nfnte vale at the center of the vortex. Dervaton of the Vortct Eqaton We wll now derve an eqaton for the vortct. We start wth the dmensonless momentm eqatons. We have t + j = p + Re We slghtl modf ths eqaton b ntrodcng the followng alternatve form of the non-lnear term j = ) j j + ɛ jk ω j k ths can readl be derved sng tensor manplaton. We take the crl ɛ pq x q ) of the momentm eqatons and fnd t ɛ pq x q ) ) + ɛ pq x q p j j + ɛ pq ɛ jk ω j k ) = ɛ pq + x q x q Re ɛ pq x q The second and the forth terms are zero, snce ɛ pq s ant-smmetrc n, q and x q s smmetrc n, q ɛ pq x q = The frst term and the last term can be wrtten as dervatves of the vortct, and we now have the followng from of the vortct eqaton ω p t + ɛ pq ɛ jk ω j k ) = x q Re ω p The second term n ths eqaton can be wrtten as ɛ pq ɛ jk ω j k ) = δ pj δ qk δ pk δ qj ) ω j k ) = x q x q k ω p x k + ω p k x k p ω j ω j p x k ω p k ) De to contnt ωj = ɛ q jpq x p = and we are left wth the relaton Ths, the vortct eqaton becomes ω p p ɛ pq ɛ jk ω j k ) = k ω j x q x k ω p t + ω p p k = ω j + x k Re ω p ω j p ) = or wrtten n vector-notaton Dω Dt = ω ) + Re ω

35 Components n Cartesan coordnates In cartesan coordnates the components of the vortct vector n three-dmensons become e x e e z ω = w x z v w = v ) e x + z z w ) v e + x x ) e z For two-dmensonal flow t s eas to see that ths redces to a non-zero vortct n the spanwse drecton, bt zero n the other two drectons. We have ω x =, ω =, ω z = v x Ths smplfes the nterpretaton and se of the concept of vortct greatl n two-dmensonal flow. Vortct and vscoct There exsts a sbtle relatonshp between flows wth vortct and flows on whch vscos forces pla a role. It s possble to show that ω vscos forces Ths means that wthot vscos forces there cannot be a vortct feld n the flow whch vares wth the coordnate drectons. Ths s expressed b the followng relatonshp τ j Ths can be derved n the followng manner. We have = µ = µɛ jk ω k ω k µɛ jk = µɛ jk ɛ kmn n x m n = µδ m δ jn δ n δ jm ) x m = µ = µ Terms n the two-dmensonal vortct eqaton In two dmensons the eqaton for the onl non-zero vortct component can be wrtten ω 3 t + ω 3 ω 3 + = ν ω 3 x x We wll take two examples elcdatng the meanng of the terms n the vortct eqaton. Frst the dffson of vortct and second the advecton or transport of vortct. Example. Dffson of vortct: the Stokes st problem revsted. From the solton of Stokes st problem above, we fnd that the vortct has the followng form ω 3 = ω z = ω, t) and that t ths s a solton of the followng dffson eqaton ω t = ν ω The tme evolton of the vortc can be seen n fgre.7, and s gven b ω = πνt e η, where η = νt where the factor νt can be dentfed as the dffson length scale, where we have sed the followng dmensonal relatons [ν] = L T, [t] = T [ νt] = L

36 ee vectorfelds // to bondar L h x x p > p p n ds n t x p η = 4νt = η 4νt t η ω z = ωη) πνt t p η = 4νt = η 4νt t η ω z = ωη) πνt t t Fgre.7: A sketch of the evolton of vortct n Stokes st problem. ω v, Fgre.8: Stagnaton pont flow wth a vscos laer close to the wall. {}}{ ω x, In ths flow the streamlnes are straght lnes and therefore the non-lnear term, or the transport/advecton term, s zero. We now trn to an example where the transport of vortct s mportant. Example 3. Transport/advecton of vortct: Stagnaton pont flow. The stagnaton pont flow s depcted n fgre.8, and s a flow where a nform veloct approaches a plate where t s showed down and trned to a flow parallel two the plates n both the postve and negatve x-drecton. At the orgn we have a stagnaton pont. The nvscd stagnaton pont flow s gven as follows nvscd: { = cx v = c ω z = ω = v x = Note that ths flow has zero vortct and that t does not satsf the no-slp condton = v =, = ). In order to satsf ths condton we ntrodce vscost, or eqall, vortct. Ths close to the plate we have to flfll the eqaton ω x + v ω ) = ν ω x + ω The two frst terms represent advecton or transport of vortct. We wll tr a smple modfcaton of the nvscd flow close to the bondar, whch s wrtten as { = xf ) v = f) Ths satsfes the contnt eqaton, and wll be zero on the bondar and approach the nvscd flow far above the plate f we appl the followng bondar condtons { = v =, = cx, v c, The vortct can be wrtten ω = xf ). We ntrodce ths and the above defntons of the veloct components nto the vortct eqaton, and fnd f f + ff + νf =

37 p η = 4νt = η 4νt t η ω z = F, F, F ωη) πνt t.5 F F.5 η F η Fgre.9: Solton to the non-dmensonal eqaton for stagnaton pont flow. F s proportonal to the normal veloct, F to the streamwse veloct and F to the vortct. The bondar condtons become { f = f = = f c, f c, An ntegral to the eqaton exsts, whch can be wrtten f ff νf = K = {evalate } = c Ths s as far as we can come analtcall and to make frther nmercal treatment smpler we make the eqaton non-dmensonal wth ν and c. Note that the have the dmensons [ν] = L T, [c] = T and that we can se the length scale ν/c and the veloct scale νc to scale the ndependent and the dependent varables as η = ν/c F η) = f) νc Ths reslts n the followng non-dmensonal eqaton F F F F = wth the bondar condtons { F = F =, η = F η, η In fgre.9 a solton of the eqaton s shown. It can be seen that dffson of vortct from the wall s balanced b transport downward b v and otward b. The thn laer of vortcal flow close to the wall has the thckness ν/c. Stretchng and tltng of vortex lnes In three-dmensons the vortct s not restrcted to a sngle non-zero component and the three-dmensonal vortct eqaton governng the vortct vector becomes ω t + ω j }{{} advecton of vortct = ω j + Re ω }{{} dffson of vortct

38 η = 4νt = η 4νt t η ω z = ωη) πνt t dr Ct) Fgre.: A closed materal crve. The complex moton of the vortct vector for a fll three-dmensonal flow can be nderstood b an analog to the eqaton governng the evolton of a materal lne, dscssed n the frst chapter. The eqaton for a materal lne s D Dt dr = dr j Note that f we dsregard the dffson term,.e. the last term of the vortct eqaton, the two eqatons are dentcal. Ths we can draw the followng conclsons abot the development of the vortex vector ) Stretchng of vortex lnes lne ω) prodce ω lke stretchng of dr prodces length ) Tltng vortex lnes prodce ω n one drecton at expense of ω n other drecton Helmholz and Kelvn s theorems The analog wth the eqaton for the materal lne drectl gve s a proof of Helmholz theorem. We have Theorem. Helmholz theorem for nvscd flow: Vortex lnes are materal lnes n nvscd flow. A second theorem vald n nvscd flow s Kelvn s theorem. Theorem. Kelvn s theorem for nvscd flow: Crclaton arond a materal crve s constant n nvscd flow. Proof. We show ths b consderng the crclaton for a closed materal crve, see fgre., Γ m = Ct) dr We can evalate the materal tme dervatve of ths expresson wth the se of Lagrangan coordnates as follows DΓ m = D dr Dt Dt Ct) Lagrange coordnates = = x, ˆt) = x m), ˆt) r = r x, ˆt) = r x m), ˆt) = d dˆt = m m ˆt r m dm The last term of ths expresson wll vansh snce m ˆt r m dm + ) r dm = m m m dm = m m ˆt ) r dm m ) m dm =

39 = η, v 4νt t η ω z = ωη) πνt t =, v) d dx x, Fgre.: A streamlne whch s tangent to the veloct vector. Ths we fnd the followng resltng expresson DΓ m Dt D = Ct) Dt dr [ = p + ] Ct) Re dr = dr, Re Re Whch shows that the crclaton for a materal crve changes onl f the vscos force Ct) Streamfncton We end ths secton wth the defnton and some propertes of the streamfncton. To defne the streamfncton we need the streamlnes whch are tangent to the veloct vector. In two dmensons, see fgre., we have dx d = v dx = d v d v dx = and n three dmensons we can wrte dx = d v = dz w For two-dmensonal flow we can fnd the streamlnes sng a streamfncton. We can ntrodce a streamfncton ψ = ψx, ) wth the propert that the streamfncton s constant along streamlnes. We can make the followng dervaton dψ = ψ x = dx + ψ d { assme = ψ = d v dx =, on streamlnes, v = ψ x } and check for consstenc We can also qckl check that the defnton gven above satsfes contnt. We have x + v = ψ x ψ x = Another propert of the streamfncton s that the volme flx between two streamlnes can be fond from the dfference b ther respectve streamfnctons. We se fgre. and fgre.3 n the followng dervatons of the volme flx:

40 ω z = πνt nt ds n t x p η = 4νt = η 4νt t η ω z = ωη) πνt t d B S n A Fgre.: Integraton paths between two streamlnes. ds n x Ψ = Ψ B Ψ = Ψ A n n = dx Fgre.3: The normal vector to a lne element. Q AB = = = = B A B n ds n x + n v) ds A { ds d =, ds n x dx = } n B d v dx) = B A A dψ = ψ B ψ A.3 Potental flow Flows whch can be fond from the gradent of a scalar fncton, a so called potental, wll be dealt wth n ths secton. Workng wth the veloct potental, for example, wll greatl smplf calclatons, bt also restrct the valdt of the soltons. We start wth a defnton of the veloct potental, then dscss smplfcatons of the momentm eqatons when sch potentals exst, and fnall we deal wth two-dmensonal potental flows, where analtc fncton theor can be sed. Veloct potental Assme that the veloct can be fond from a potental. In three dmensons we have = φ or = φ wth the correspondng relatons n two dmensons = φ x v = φ There are several conseqences of ths defnton of the veloct feld.

41 ) Potental flow have no vortct, whch can be seen b takng the crl of the gradent of the veloct potental,.e. k φ ω = ɛ jk = ɛ jk = x k ) Potental flow s not nflenced b vscos forces, whch s seen b the relatonshp between vscos forces and vortct, derved earler. We have τ j = µ = µɛ jk ω k ) Potental flow satsfes Laplace eqaton, whch s a conseqence of the dvergence constrant appled to the gradent of the veloct potental,.e. Bernoll s eqaton = φ = φ = For potental flow the veloct feld can ths be fond b solvng Laplace eqaton for the veloct potental. The momentm eqatons can then be sed to fnd the pressre from the so called Bernoll s eqaton. We start b rewrtng the momentm eqaton t + j = p + ν gδ 3 ρ sng the alternatve form of the non-lnear terms,.e. and obtan the followng eqaton t + j = ) k k ɛ jk j ω k k k + p ) ρ + gx ω k 3 = ɛ jk j ω k + νɛ jk Ths eqaton can be ntegrated n for two dfferent cases: potental flow wthot vortct) and statonar nvscd flow wth vortct. ) Potental flow. Introdce the defnton of the veloct potental = φ above, we have φ t + k k + p ) ρ + gx 3 = φ t ) + p ρ + gx 3 = ft) nto the momentm eqaton ) Statonar, nvscd flow wth vortct. Integrate the momentm eqatons along a streamlne, see fgre.4. We have t k k + p ) ρ + gx 3 ds = ɛ jk jω k ds = + + 3) + p ρ + gx 3 = constant along streamlnes Ths, Bernoll s eqaton can be sed to calclate the pressre when the veloct s known, for two dfferent flow statons. Example 4. Ideal vortex: Swrlng statonar flow wthot vortct. The veloct potental = φ n polar coordnates can be wrtten r = φ r θ = r φ θ

42 h t x x η p > p ω z = ωη) πνt p t n t = ds n t Fgre.4: Integraton x of the momentm eqaton along a streamlne. p η = 4νt = η 4νt t η ω z = ωη) πνt t r Fgre.5: Streamlnes for the deal vortex. The veloct potental satsfes Laplace eqaton, whch n polar coordnates becomes φ = ) φ + φ r r r r r θ = The frst term vanshes snce φ r = r =. Integrate the resltng eqaton twce and we fnd φ = Cθ + D The constant D s arbtrar so we can set t to zero. Usng the defnton of the crclaton we fnd θ = C r, C = Γ π We can easl fnd the streamfncton for ths flow. In polar coordnates the streamfncton s defned r = ψ r θ θ = ψ r Note that the streamfncton satsfes contnt,.e. =. We se the solton from the veloct potental to fnd the streamfncton, we have ψ r = Γ πr ψ = Γ π ln r The streamfncton s constant along streamlnes resltng n crclar streamlnes, see fgre.5 We can now se Bernoll s eqaton to calclate the pressre dstrbton for the deal vortex as θ + p = p p = p Γ 8π r Note that the solton has a snglart at the orgn. Two-dmensonal potental flow sng analtc fnctons For two-dmensonal flows we can se analtc fncton theor,.e. complex valed fnctons, to calclate the veloct potental. In fact, as we wll show n the followng. An analtc fncton represents a twodmensonal potental flow.

43 We start b showng that both the veloct potental and the streamfncton satsf Laplace eqaton. Veloct potental φ for two-dmensonal flow s defned = φ x Usng the contnt eqaton gves The streamfncton ψ s defned v = φ x + v = φ = = ψ v = ψ x B notng that potental flow has no vortct we fnd ω = v x + = ψ = Ths, both the veloct potental φ and the streamfncton ψ satsf Laplace eqaton. We can tlze analtc fncton theor b ntrodcng the complex fncton where F z) = φx, ) + ψx, ) z = x + = re θ = rcos θ + sn θ) whch wll be denoted the complex potental. We now recall some sefl reslts from analtc fncton theor. F z) exsts and s nqe F z) s an analtc fncton The fact that a dervatve of a complex fncton shold be nqe rests on the valdt of the Cach Remann eqatons. We can derve those eqatons b reqrng that a dervatve of a complex fncton shold be the same ndependent of the drecton we take the lmt n the complex plane, see fgre.6. We have F z) = lm z = lm x = lm [F z + z) F z)] z [φx + x, ) φx, ) + ψx + x, ) ψx, )] x [φx, + ) φx, ) + ψx, + ) ψx, )] φ x = ψ ψ x = φ ψ x = φ real part magnar part Ths we have the Cach Remann eqatons as a necessar reqrement for a complex fncton to be analtc,.e. φ x = ψ φ and = ψ x Usng the Cach-Remann eqatons we can show that the real φ and magnar part ψ of the complex potental satsf Laplace eqatons and ths that the are canddates for dentfcaton wth the veloct potental and streamfncton of a two-dmensonal potental flow. We have φ x = ) ψ = ) ψ = φ x x ψ = ) φ = x x ) φ = ψ x

44 p η = 4νt = η 4νt t η ω z = ωη) πνt t x }{{} }{{} x Fgre.6: The complex plane and the defntons of analtc fnctons. In addton to satsfng the Laplace eqatons, we also have to show that the Cach-Remann eqatons are consstent wth the defnton of the veloct potental and the streamfncton. The Cach Remann eqatons are = φ x = ψ v = φ = ψ x whch recovers the defntons of the veloct potental and the streamfncton. Ths an analtc complex fncton can be dentfed wth a potental flow, where the real part s the veloct potental and the magnar part s the streamfncton. We can now defne the complex veloct as the complex conjgate of the veloct vector snce W z) = df dz = φ x + ψ x = v It wll be sefl to have a smlar relaton for polar coordnates, see fgre.7. In ths case we have { = r cos θ θ sn θ v = r sn θ + θ cos θ W z) = v = [ r r, θ) θ r, θ)] e θ }{{} cos θ sn θ We wll now take several examples of how potental veloct felds can be fond from analtc fnctons. Example 5. Calclate the veloct feld from the complex potental F = Ue α z. The complex veloct becomes W z) = F = Ue α = U cos α sn α whch gves s the solton for the veloct feld n phscal varables, see fgre.8, as { = U cos α v = U sn α Example 6. Calclate the veloct feld from the complex potental F = Az n. In polar coordnates the complex potental becomes Ar n e nθ = Ar n cosnθ) + Ar n snnθ) Ths the veloct potental and the streamfncton can be dentfed as { φ = Ar n cosnθ) ψ = Ar n snnθ)